We have $A,\ B,\ C\subset \mathbb{R}^{p}$, nonempty, $C$ is bounded and $A+C \subset B+C$. Prove that $A \subset {\rm cl}({\rm conv}B)$.
For ${\rm cl}({\rm conv}B)$, I started using the definition with sequences of a closed set, but I don't know how to reffer to the information that $C$ is bounded.
Here we need the condition where $C$ is bounded : If not, assume that $C = \mathbb{R}^p$. Hence for any $A,\ B$ we have $A+C\subset B+C$
Assume that $C$ is bounded and $A+C\subset B+C$ Hence $A+C \subset X+ C$ where $X ={\rm cl\ conv} \ B$. Assume that $a\in A-X$. Hence there is a plane $l$ s.t. $l$ is between $a$ and $X$, since $X$ is convex.
Here we can assume that origin $o$ is in $C$, without generality.
There is $v\in C$ s.t. $v$ is orthogonal to $l$ and $|v|$ is largest. Hence $a+v \in X+C$, which is a contradiction. Hence $A\subset X$.